The most important way to characterize a random variable is through the probabilities of the values that it can take. For a discrete random variable XXX, these are captured by the probability mass function (PMF for short) of XXX, denoted pXp_XpX. In particular, for any real number xxx, the probability mass of xxx. denoted pX(x)p_X(x)pX(x). is the probability of the event {X=x}\{X = x\}{X=x}. Thus, from the additivity and normalization axioms, we have
∑xpX(x)=1\sum_{x}p_X(x)=1x∑pX(x)=1

In what follows, we will often omit the braces from the event/set notation when no ambiguity can arise. In particular, we will usually write P(X=x)P(X = x)P(X=x) in place of the more correct notation P({X=x})P(\{X = x\})P({X=x}).

We will use upper case characters to denote random variables, and lower case characters to denote real numbers such as the numerical values of a random variable.

The Bernoulli Random Variable (伯努利随机变量)

Consider the toss of a coin, which comes up a head with probability ppp, and a tail with probability 1−p1 - p1−p. The Bernoulli random variable takes the two values 1 and 0, depending on whether the outcome is a head or a tail:

Its PMF is

The Binomial Random Variable

A coin is tossed nnn times. At each toss, the coin comes up a head with probability ppp, and a tail with probability 1−p1 - p1−p, independent of prior tosses. Let XXX be the number of heads in the nnn-toss sequence. We refer to XXX as a binomial random variable with parameters nnn and ppp. The PMF of XXX consists of the binomial probabilities:
pX(k)=P(X=k)=(nk)pk(1−p)n−k,k=0,1,...,n.p_X(k) = P(X = k) =\begin{pmatrix}n\\k \end{pmatrix}p^k(1-p)^{n-k},k = 0, 1, ... , n.pX(k)=P(X=k)=(nk)pk(1−p)n−k,k=0,1,...,n.The normalization property, specialized to the binomial random variable, is written as
∑k=0n(nk)pk(1−p)n−k=1\sum_{k=0}^n\begin{pmatrix}n\\k \end{pmatrix}p^k(1-p)^{n-k}=1k=0∑n(nk)pk(1−p)n−k=1

Form of the binomial PMF.

Let k∗=⌊(n+1)p⌋k^*=\lfloor (n + 1)p\rfloork∗=⌊(n+1)p⌋. The PMF pX(k)p_X(k)pX(k) is monotonically nondecreasing with kkk in the range from 000 to k∗k^*k∗. and is monotonically decreasing with kkk for k≥k∗k\geq k^*k≥k∗.
(pX(k)pX(k−1)=(n+1)p−kpk−kp)(\frac{p_X(k)}{p_X(k-1)}=\frac{(n+1)p-kp}{k-kp})(pX(k−1)pX(k)=k−kp(n+1)p−kp)

Problem 6.

The Celtics and the Lakers are set to play a playoff series of nnn basketball games, where nnn is odd. The Celtics have a probability ppp of winning any one game, independent of other games. For any k>0k > 0k>0, find the values for ppp for which n=2k+1n = 2k + 1n=2k+1 is better for the Celtics than n=2k−1n = 2k -1n=2k−1.

SOLUTION

Let NNN be the number of Celtics’ wins in the first 2k−12k -12k−1 games. If AAA denotes the event that the Celtics win with n=2k+1n = 2k +1n=2k+1, and BBB denotes the event that the Celtics win with n=2k−1n = 2k-1n=2k−1, then
P(A)=P(N≥k+1)+P(N=k)⋅(1−(1−p)2)+P(N=k−1)⋅p2P(B)=P(N≥k)=P(N=k)+P(N≥k+1)P(A)=P(N\geq k+1)+P(N=k)\cdot(1-(1-p)^2)+P(N=k-1)\cdot p^2\\ P(B)=P(N\geq k)=P(N=k)+P(N\geq k+1)P(A)=P(N≥k+1)+P(N=k)⋅(1−(1−p)2)+P(N=k−1)⋅p2P(B)=P(N≥k)=P(N=k)+P(N≥k+1)and therefore
P(A)−P(B)=P(N=k−1)⋅p2−P(N=k)⋅(1−p)2=(2k−1)!(k−1)!k!pk(1−p)k(2p−1)\begin{aligned}P(A)-P(B)&=P(N=k-1)\cdot p^2-P(N=k)\cdot (1-p)^2\\&=\frac{(2k-1)!}{(k-1)!k!}p^k(1-p)^k(2p-1)\end{aligned}P(A)−P(B)=P(N=k−1)⋅p2−P(N=k)⋅(1−p)2=(k−1)!k!(2k−1)!pk(1−p)k(2p−1)It follows that P(A)>P(B)P(A) > P(B)P(A)>P(B) if and only if p>12p > \frac{1}{2}p>21. Thus, a longer series is better for the better team.

The Geometric Random Variable

几何随机变量

Suppose that we repeatedly and independently toss a coin with probability of a head equal to ppp, where 0<p<10 < p < 10<p<1. The geometric random variable is the number XXX of tosses needed for a head to come up for the first time. Its PMF is given by
pX(k)=(1−p)k−1p,k=1,2,...,p_X(k)=(1-p)^{k-1}p,k=1,2,...,pX(k)=(1−p)k−1p,k=1,2,...,
More generally, we can interpret the geometric random variable in terms of repeated independent trials until the first “success.”

The Poisson Random Variable

泊松随机变量

A Poisson random variable has a PMF given by
pX(k)=e−λλkk!,k=0,1,2,...,p_X(k)=e^{-\lambda}\frac{\lambda^k}{k!},\ \ \ \ \ \ \ \ k=0,1,2,...,pX(k)=e−λk!λk, k=0,1,2,...,where λ\lambdaλ is a positive parameter characterizing the PMF. This is a legitimate PMF because
∑k=0∞e−λλkk!=e−λeλ=1\sum_{k=0}^\infty e^{-\lambda}\frac{\lambda^k}{k!}=e^{-\lambda}e^{\lambda}=1k=0∑∞e−λk!λk=e−λeλ=1

Form of the Poisson PMF.

The PMF pX(k)p_X(k)pX(k) increases monotonically with kkk up to the point where kkk reaches the largest integer not exceeding λ\lambdaλ, and after that point decreases monotonically with kkk.
(pX(k)pX(k−1)=λk)(\frac{p_X(k)}{p_X(k-1)}=\frac{\lambda}{k})(pX(k−1)pX(k)=kλ)

Poisson approximation property

The Poisson PMF with parameter λ\lambdaλ is a good approximation for a binomial PMF with parameters nnn and ppp. i.e …
e−λλkk!≈n!k!(n−k)!pk(1−p)n−k,ifk≪ne^{-\lambda}\frac{\lambda^k}{k!}\approx \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k},\ \ \ \ \ \ \ \ if\ k\ll ne−λk!λk≈k!(n−k)!n!pk(1−p)n−k, if k≪nprovided λ=np\boldsymbol{\lambda= np}λ=np. nnn is very large, and ppp is very small. In this case. using the Poisson PMF may result in simpler models and calculations.
- For example. let n=100n = 100n=100 and p=0.01p = 0.01p=0.01. Then the probability of k=5k = 5k=5 successes in n=100n = 100n=100 trials is calculated using the binomial PMF as 0.002900.002900.00290. Using the Poisson PMF with λ=np=100⋅0.01=1\lambda= np = 100\cdot0.01 = 1λ=np=100⋅0.01=1. this probability is approximated by 0.003060.003060.00306.
Proof: Consider the PMF of a binomial random variable with parameters n→∞n\rightarrow\inftyn→∞ and p→0p\rightarrow0p→0 while npnpnp is fixed at a given value λ\lambdaλ
pX(k)=n!(n−k)!k!pk(1−p)n−k=n(n−1)...(n−k+1)nkλkk!(1−λn)n−kn−k+jn→1,(1−λn)k→1,(1−λn)n→e−λp_X(k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}=\frac{n(n-1)...(n-k+1)}{n^k}\frac{\lambda^k}{k!}(1-\frac{\lambda}{n})^{n-k}\\ \frac{n-k+j}{n}\rightarrow1,(1-\frac{\lambda}{n})^{k}\rightarrow1,(1-\frac{\lambda}{n})^{n}\rightarrow e^{-\lambda}pX(k)=(n−k)!k!n!pk(1−p)n−k=nkn(n−1)...(n−k+1)k!λk(1−nλ)n−knn−k+j→1,(1−nλ)k→1,(1−nλ)n→e−λThus, for each fixed kkk, as n→∞n\rightarrow\inftyn→∞ we obtain
pX(k)→e−λλkk!p_X(k)\rightarrow e^{-\lambda}\frac{\lambda^k}{k!}pX(k)→e−λk!λk

Functions of Random Variables

Given a random variable XXX, one may generate other random variables by applying various transformations on XXX. If Y=g(X)Y = g(X)Y=g(X) is a function of a random variable XXX, then YYY is also a random variable, since it provides a numerical value for each possible outcome.
If XXX is discrete with PMF pXp_XpX. then YYY is also discrete, and its PMF pYp_YpY can be calculated using the PMF of XXX.
pY(y)=∑{x∣g(x)=y}pX(x)p_Y(y)=\sum_{\{x|g(x)=y\}}p_X(x)pY(y)={x∣g(x)=y}∑pX(x)

References

IntroductionIntroductionIntroduction tototo ProbabilityProbabilityProbability

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Chapter 2 (Discrete Random Variables): Probability mass functions (PMF 分布列)

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